2. ## Re: Geometric progression

Originally Posted by parmis

I'm going to take a guess here. Is this your problem?

$a_n = \frac{(m^2 - a)n^2 +m + n}{m^2 + 2m + 1}$

But that can't be right...the a on the RHS needs a subscript, doesn't it?

-Dan

3. ## Re: Geometric progression

Oh... sorry i made a mistake. a n is Arithmetic progression! d=1/5

4. ## Re: Geometric progression

So $a_n$ is an arithmetic progression of the form $a_n= a+ nr$ but what does "m" represent in your formula?

5. ## Re: Geometric progression

I dont know. My teacher gave me this, but i cant answer it!

6. ## Re: Geometric progression

Then you should probably ask your teacher what $m$ means.

7. ## Re: Geometric progression

Originally Posted by parmis
Oh... sorry i made a mistake. a n is Arithmetic progression! d=1/5
Assuming your arithmetic progression is $a_n = a+dn$ since you arbitrarily wrote a d for some reason, we can set these two expressions equal to each other:

$a_n = a+dn$

$= a+\dfrac{n}{5} = \dfrac{(m^2-a)n^2+m+n}{m^2+2m+1}$

If you are trying to solve for $m$, it is the solution to this quadratic equation in $m$:

$\left(a+\dfrac{n}{5}-n^2\right)m^2 + \left(2a + \dfrac{2n}{5}-1\right)m + \left(a+an^2-\dfrac{4n}{5}\right) = 0$

You can use the quadratic formula. You get a mess, but I suppose that could be the solution for $m$...